Question: If $x \diamond y = 4x^{2}+y^{2}$ and $x \triangleleft y = (4-x)(y)$, find $4 \triangleleft (-5 \diamond -6)$.
Solution: First, find $-5 \diamond -6$ $ -5 \diamond -6 = 4(-5)^{2}+(-6)^{2}$ $ \hphantom{-5 \diamond -6} = 136$ Now, find $4 \triangleleft 136$ $ 4 \triangleleft 136 = (4-4)(136)$ $ \hphantom{4 \triangleleft 136} = 0$.